University Calculus: Early Transcendentals (3rd Edition)

$$\frac{dy}{dx}=\frac{1}{y^{-1/2}+1}$$ $$\frac{d^2y}{dx^2}=\frac{y^{-3/2}}{2(y^{-1/2}+1)^3}$$
$$2\sqrt y=x-y$$ $$2y^{1/2}=x-y$$ 1) Find $dy/dx$: Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(2y^{1/2})=\frac{d}{dx}(x)-\frac{d}{dx}(y)$$ $$2\times\frac{1}{2}y^{-1/2}\frac{dy}{dx}=1-\frac{dy}{dx}$$ $$y^{-1/2}\frac{dy}{dx}=1-\frac{dy}{dx}$$ Isolate the terms with $dy/dx$ into one side and solve for $dy/dx$: $$y^{-1/2}\frac{dy}{dx}+\frac{dy}{dx}=1$$ $$(y^{-1/2}+1)\frac{dy}{dx}=1$$ $$\frac{dy}{dx}=\frac{1}{y^{-1/2}+1}$$ 2) Find $d^2y/dx^2$: Differentiate the derivative $dy/dx$ as normal, using Quotient Rule: $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{1}{y^{-1/2}+1}\Big)=\frac{-1\frac{d}{dx}(y^{-1/2}+1)}{(y^{-1/2}+1)^2}=\frac{-\Big(-\frac{1}{2}y^{-3/2}\frac{dy}{dx}\Big)}{(y^{-1/2}+1)^2}$$ $$\frac{d^2y}{dx^2}=\frac{y^{-3/2}\frac{dy}{dx}}{2(y^{-1/2}+1)^2}$$ Now we can use the result $dy/dx=1/(y^{-1/2}+1)$ from part 1) to do substitution here: $$\frac{d^2y}{dx^2}=\frac{\frac{y^{-3/2}}{y^{-1/2}+1}}{2(y^{-1/2}+1)^2}=\frac{y^{-3/2}}{2(y^{-1/2}+1)^3}$$